In $L_1$, the number of $a$'s are more than double of $b$'s, so there has to be atleast one $a$ (when there are no $b$'s). Also, for every addition of $b$, atleast 2 $a$'s must be added. You can generate $L_1$ as:

$
S_1 \rightarrow aA \\
A \rightarrow aaAb \mid aA \mid \varepsilon
$

$L_2$ is a little more tricky. The number of $a$'s are less than double the number of $b$'s, so there can be $0$ $a$'s, but non-zero $b$'s. Consider the "base case" of $1$ $b$. The string can be either $b$ or $ab$. We will let the first rule generate this base case. After that, notice that for every addition of $b$, we can increase the number of $a$'s by at most $2$. So we can add $0$, $1$, or $2$ $a$'s for every addition of $b$. We'll let the second rule handle this. So, the CFG for $L_2$ becomes:

Note that CFG's are closed under union, that is, the union of two CFG's is also a CFG. So, to get the CFG for $L$, let the starting state $S$ of $L$ lead to either the starting state of $L_1$, or of $L_2$:

$S \rightarrow S_1 \mid S_2$

The rest of the rules remain the same as in the two languages. There may be a simpler grammar, but this was the first that came to mind.

Start with a grammar for $\{a^i b^{2 i} : i \ge 0\}$:
$$
S \rightarrow a S bb \mid \epsilon
$$
Now hack it so that there are more $a$ than $bb$:
$$
\begin{align*}
S &\rightarrow a S bb \mid A \\
A &\rightarrow a A \mid a
\end{align*}
$$
The modificaton to force more $b$ is similar, and left to you. It adds a nonterminal and three productions.